Question

cos (A - B) = cos A cos B + sin A sin B
If true then write 1 and if false then write 0

Answer 1

Given, $A={60}^{\circ }$ and $B={30}^{\circ }$
Now, $\cos (A-B)=\cos A\cos B+\sin A\sin B$
L.H.S : $\cos (A-B)=\cos ({60}^0-{30}^0)=\cos {30}^0=\frac{\sqrt{3}}{2}$
R.H.S: $\cos A\cos B+\sin A\sin B=\cos {60}^0\cos {30}^0+\sin {60}^0\sin {30}^0$
= $\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)$
= $\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}$
= $\frac{\sqrt{3}}{2}$

$\therefore LHS=RHS$
$\therefore \cos (A-B)=\cos A\cos B+\sin A\sin B$